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Part 1A Simple Approach

ToShort Circuit

Calculations

Bulletin EDP-1(2004-1)

EngineeringDependableProtectionFor AnElectricalDistributionSystem

Bussmann

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Electrical Distribution System

Basic Considerations of Short-Circuit Calculations

Sources of short circuit current that are normally taken

under consideration include:

- Utility Generation

- Local Generation

- Synchronous Motors and- Induction Motors

Capacitor discharge currents can normally be

neglected due to their short time duration. Certain IEEE

(Institute of Electrical and Electronic Engineers) publications

detail how to calculate these currents if they are substantial.

Asymmetrical ComponentsShort circuit current normally takes on an asymmetrical

characteristic during the first few cycles of duration. That is,

it is offset about the zero axis, as indicated in Figure 1.

Figure 1

In Figure 2, note that the total short circuit current Ia is

the summation of two components - the symmetrical RMS

current IS, and the DC component, IDC. The DC componentis a function of the stored energy within the system at the

initiation of the short circuit. It decays to zero after a few

cycles due to I2R losses in the system, at which point the

short circuit current is symmetrical about the zero axis. The

RMS value of the symmetrical component may be deter-

mined using Ohm`s Law. To determine the asymmetrical

component, it is necessary to know the X/R ratio of the

system. To obtain the X/R ratio, the total resistance and total

reactance of the circuit to the point of fault must be

determined. Maximum thermal and mechanical stress on

the equipment occurs during these first few cycles. It is

important to concentrate on what happens dur ing the first

half cycle after the initiation of the fault.

3

Why Short-Circuit CalculationsSeveral sections of the National Electrical Code relate

to proper overcurrent protection. Safe and reliable

application of overcurrent protective devices based on

these sections mandate that a short circuit study and aselective coordination study be conducted.

These sections include, among others:

110-9 Interrupting Rating

110-10 Component Protection

230-65 Service Entrance Equipment

240-1 Conductor Protection

250-95 Equipment Grounding Conductor Protection

517-17 Health Care Facilities - Selective Coordination

Compliance with these code sections can best be

accomplished by conducting a shor t circuit study and a

selective coordination study.

The protection for an electrical system should not onlybe safe under all service conditions but, to insure continuity

of service, it should be selectively coordinated as well. A

coordinated system is one where only the faulted circuit is

isolated without disturbing any other part of the system.

Overcurrent protection devices should also provide short-

circuit as well as overload protection for system

components, such as bus, wire, motor controllers, etc.

To obtain reliable, coordinated operation and assure

that system components are protected from damage, it is

necessary to first calculate the available fault current at

various critical points in the electrical system.

Once the short-circuit levels are determined, the

engineer can specify proper interrupting rating require-

ments, selectively coordinate the system and providecomponent protection.

General Comments on Short-Circuit CalculationsShort Circuit Calculations should be done at all critical

points in the system.

These would include:

- Service Entrance

- Panel Boards

- Motor Control Centers

- Motor Starters

- Transfer Switches

Normally, short circuit studies involve calculating abolted 3-phase fault condition. This can be characterized

as all three phases bolted together to create a zero

impedance connection. This establishes a worst case

condition, that results in maximum thermal and mechanical

stress in the system. From this calculation, other types of

fault conditions can be obtained.

C

U

R

R

E

N

T

TIME

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3 Short-Circuit Current Calculations

Procedures and Methods

3 Short-Circuit Current Calculations,Procedures and Methods

To determine the fault current at any point in the

system, first draw a one-line diagram showing all of the

sources of short-circuit current feeding into the fault, as well

as the impedances of the circuit components.To begin the study, the system components, including

those of the utility system, are represented as impedances

in the diagram.

The impedance tables given in the Data Section

include three phase and single phase transformers, current

transformers, safety switches, circuit breakers, cable, and

busway. These tables can be used if information from the

It must be understood that short circuit calculations are

performed without current limiting devices in the system.

Calculations are done as though these devices are

replaced with copper bars, to determine the maximum

available short circuit current. This is necessary to

project how the system and the current limiting devices willperform.

Also, current limiting devices do not operate in series

to produce a compounding current limiting effect. The

downstream, or load side, fuse will operate alone under a

short circuit condition if properly coordinated.

5

System A3 Single Transformer System

M

Available UtilityS.C. MVA 100,000

1500 KVA Transformer480Y/277V,3.5%Z, 3.45%X, .56%RIf.l. = 1804A

25 - 500kcmil6 Per PhaseService Entrance Conductorsin Steel Conduit

2000A Switch

KRP-C-2000SP FuseMain Swbd.

400A Switch

LPS-RK-400SP Fuse

50 - 500 kcmilFeeder Cable in Steel Conduit

Fault X2MCC No. 1

Motor

Fault X1

System B3 Double Transformer System

1000 KVA Transformer,

480/277 Volts 33.45%X, .60%RIf.l. = 1203A

Available UtilityS.C. KVA 500,000

30 - 500 kcmil4 Per Phase

Copper in PVC Conduit 1600A Switch

KRP-C-1500SP Fuse

LPS-RK-350SP Fuse

400A Switch

225 KVA208/120 Volts 3.998%X, .666%R

20 - 2/02 Per PhaseCopper in PVC Conduit

In this example, assume 0% motor load.

2

To begin the analysis, consider the following system,

supplied by a 1500 KVA, three phase transformer having a

full load current of 1804 amperes at 480 volts. (See System

A, below) Also, System B, for a double transformation, will

be studied.To start, obtain the available short-circuit KVA, MVA, or

SCA from the local utility company.

The utility estimates that System A can deliver a short-

circuit of 100,000 MVA at the primary of the transformer.

System B can deliver a short-circuit of 500,000 KVA at the

primary of the first transformer. Since the X /R ratio of the

utility system is usually quite high, only the reactance need

be considered.

With this available short-circuit information, begin to

make the necessary calculations to determine the fault

current at any point in the electrical system.

Four basic methods will be presented in this text to

instruct the reader on short circuit calculations.

These include :- the ohmic method

- the per unit method

- the TRON Computer Software method

- the point to point method

1

2

Note: The above 1500KVA transformer serves 100% motor load.

Fault X1

Fault X2

1

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3 Short-Circuit Current Calculations Procedures and Methods

Ohmic Method

6

3 Short Circuit Calculations,Ohmic Method

Most circuit component impedances are given in ohms

except utility and transformer impedances which are found

by the following formulae* (Note that the transformer andutility ohms are referred to the secondary KV by squaring

the secondary voltage.)

Step 1. Xutility =1000 (KVsecondary)2

S.C. KVAutility

Step 2. Xtrans =(10)(%X**)(KVsecondary)2

KVAtrans

Rtrans =(10)(%R**)(KVsecondary)2

KVAtrans

Step 3. The impedance (in ohms) given for currenttransformers, large switches and large circuit breakers isessentially all X.

Step 4. Xcable and bus .Rcable and bus.

Step 5. Total all X and all R in system to point of fault.

Step 6. Determine impedance (in ohms) of the system by:

ZT = (RT)2 + (XT)2

Step 7. Calculate short-circuit symmetrical RMS amperes

at the point of fault.

IS.C. sym RMS=Esecondary line-line

3 (ZT)

generally a percentage of the transformer full load current,

depending upon the types of loads. The generally

accepted procedure assumes 50% motor load when both

motor and lighting loads are considered, such as supplied

by 4 wire, 208Y/120V and 480Y/277V volt 3-phase

systems.)

*For simplicity of calculations all ohmic values are single phase distance one way, later compensated for in the three phase short-circuit formula by the factor, 3.(See Step 7.)

**UL Listed transformers 25 KVA and larger have a 10% impedance tolerance. Short circuit amperes can be affected by this tolerance.Only X is considered in this procedure since utility X/R ratios are usually quite high. For more finite details obtain R of utility source.A more exact determination depends upon the sub-transient reactance of the motors in question and associated circuit impedances. A less conservativemethod would involve the total motor circuit impedance to a common bus (sometimes referred to as a zero reactance bus).

Arithmetical addition results in conservative values of fault current. More finite values involve vectorial addition of the currents.

Note: The ohms of the circuit components must be referred to the same voltage. If there is more than one voltage transformation in the system, the ohmicmethod becomes more complicated. It is recommended that the per-unit method be used for ease in calculation when more than one voltage transformationexists in the system.

Step 9. The symmetrical motor contribution can beapproximated by using an average multiplying factor

associated with the motors in the system. This factor varies

according to motor design and in this text may be chosen

as 4 times motor full load current for approximatecalculation purposes. To solve for the symmetrical motor

contribution:

Isym motor contrib = (4) x (Ifull load motor)

Step 10. The total symmetrical short-circuit RMS current iscalculated as:

Itotal S.C. sym RMS = (IS.C. sym RMS ) + (Isym motor contrib)

Step 11. Determine X/R ratio of the system to the point offault.

X/Rratio =Xtotal Rtotal

Step 12. The asymmetrical factor corresponding to the X/Rratio in Step 11 is found in Table 8, Column Mm. This

multiplier will provide the worst case asymmetry occurring

in the first 1/2 cycle. When the average 3-phase multiplier

is desired use column Ma.

Step 13. Calculate the asymmetrical RMS short-circu itcurrent.

IS.C. asym RMS = (IS.C. sym RMS) x (Asym Factor)

Step 14. The short-circuit current that the motor load cancontribute is an asymmetrical current usually approximated

as being equal to the locked rotor current of the motor.

As a close approximation with a margin of safety use:

Iasym motor contrib = (5) x (Ifull load motor)

Step 15. The total asymmetrical short-circuit RMS current iscalculated as:

Itotal S.C. asym RMS = (IS.C. asym RMS) + (Iasym motor contrib)

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M M

3 Short-Circuit Current Calculations Procedures and Methods

Ohmic Method To Fault X1 System A

Available UtilityS.C. MVA 100,000

1500 KVA Transformer,480V, 3,3.5%Z, 3.45%X , 0.56%R

If.l. trans = 1804A

25 - 500 kcmil6 Per PhaseService EntranceConductors in Steel Conduit

2000A Switch

KRP-C-2000SP Fuse

Fault X1

Motor Contribution

7

R X

X =1000(.48)2

= 0.0000023 0.0000023100,000,000

X =(10) (3.45) (.48)2

= 0.0053 0.00531500

R =(10) (.56) (.48)2

= 0.00086 0.00086 1500

X =25'

x0.0379

= 0.000158 0.0001581000 6

R = 25' x 0.0244 = 0.000102 0.000102 1000 6

X = 0.000050 0.000050

Total R and X = 0.000962 0.00551

Ztotal per = (0.000962)2 + (0.00551)2 = 0.0056phase

IS.C. sym RMS =480 = 49,489A

3 (.0056)

Isym motor contrib = 4 x 1804 = 7216A(100% motor load)

Itotal S.C. sym RMS = 49,489 + 7216 = 56,705A(fault X1)

X/Rratio = .00551 = 5.73.000962

Asym Factor = 1.294 (Table 8)

IS.C. asym RMS = 1.294 x 49,489 = 64,039A

Iasym motor contrib = 5 x 1804 = 9,020A(100% motor load)

Itotal S.C. asym RMS = 64,039 + 9,020 = 73,059A(fault X1)

One-Line D iagram Impedance D iagram

1 1

Note: See Ohmic Method Procedure for Formulas.

(Table 1.2)

(Table 3)

(Table 5)

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R X

X = 0.00551 0.00551

R = 0.000962 0.000962

X = .00008 0.00008

X =50 x .0379 = 0.00189 0.00189

1000

R =50 x .0244 = 0.00122 0.00122

1000

Total R and X = 0.002182 0.00748

Ztotal per = (0.002182)2 + (0.00748)2 = 0.00778phase

IS.C. sym RMS =480 = 35,621A

3 (.00778)

Isym motor contrib = 4 x 1804 = 7216A(100% motor load)

Itotal S.C. sym RMS = 35,621 + 7,216 = 42,837A(fault X2)

X/Rratio =.00748 = 3.43

.002182

Asym Factor=

1.149 (Table 8)IS.C. asym RMS = 1.149 x 35,621 = 40,929A

Iasym motor contrib = 5 x 1804 = 9,020A(100% motor load)

Itotal S.C. asym RMS = 40,929 + 9,020 = 49,949A(fault X2)

Note: See Ohmic Method Procedure for Formulas. Actual motor contributionwill be somewhat smaller than calculated due to the impedance of thefeeder cable.

8

3 Short-Circuit Current Calculations Procedures and Methods

Ohmic Method To Fault X2 System A

M M

Fault X1

400A Switch

LPS-RK-400SP Fuse

50 - 500 kcmilFeeder Cablein Steel Conduit

Fault X2

Motor Contribution

One-Line Diagram Impedance Diagram

2 2

1 1

(Table 3)

(Table 5)

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To use the OHMIC Method through a second transformer,

the following steps apply:

Step 1a. Summarize X and R values of all components onprimary side of transformer.

One-Line Diagram Impedance Diagram

Step 1b. Reflect X and R values of all components tosecondary side of transformer

Xs =Vs2 (Xp) Rs =

Vs2 (Rp)Vp2 Vp2

and proceed with steps 2 thru 15 from page 6.

3 Short-Circuit Current Calculations Procedures and Methods

Ohmic Method To Fault X1 System B

9

R X

X =1000 (.48)2

= .000461 .000461500,000

X =(10) (3.45) (.48)2

= .00795 .007951000

R =(10) (.60) (.48)2

= .00138 .00138 1000

X =30'

x.0303 = .000227 .000227

1000 4

R =30'

x.0220

= .000165 .000165 1000 4

X = .000050 .00005

Total R and X = .001545 .008688

Ztotal per = (.001545)2 + (.008688)2 = .008824phase

IS.C. sym RMS =480 = 31,405A

3 (.008824)

X/Rratio =.008688 = 5.62.001545

Asym Factor = 1.285 (Table 8)

IS.C. asym RMS = 31,405 x 1.285 = 40,355A

Available Utility500,000 S.C. KVA

1000KVA Transformer,480V, 3,3.45% X, .60% R

30' - 500 kcmil4 Per PhaseCopper in PVC Conduit

1600A Switch

KRP-C-1500SP Fuse

1 1

(Table 1.2)

(Table 3)

(Table 5)

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3 Short-Circuit Current Calculations Procedures and Methods

Ohmic Method To Fault X2 System B

400A Switch

LPS-RK-350SP Fuse

20' - 2/02 Per PhaseCopper in PVC Conduit

225KVA Transformer,208/120V,.998%X, .666%R

R X

X = .008688 .008688

R = .001545 .001545

X = .00008 .00008

X =20' x .0327 = .000327

.000327

1000 2

R =20' x .0812 = .000812 .000812

1000 2

Total R and X (480V) = .002357 .009095

To Reflect X and R to secondary:

Xtotal =(208)2 x (.009095)

= .001708 .001708(208V) (480)2

Rtotal =(208)2 x (.002357)

= .000442 .000442 (208V) (480)2

X =(10) (.998) (.208)2

= .00192 .00192225

R =(10) (.666) (.208)2

= .00128 .00128 225

Total R and X (208V) = .001722 .003628

Ztotal per = (.001722)2 + (.003628)2 = .004015phase

IS.C. sym RMS =208 = 29,911A

3 (.004015)

X/Rratio =.003628 = 2.10.001722

Asym Factor = 1.0491 (Table 8)

IS.C. asym RMS = 29,911 x 1.0491 = 31,380A

10

One-Line Diagram Impedance Diagram

1 1

2 2

(Table 5)

(Table 1.2)

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3 Short-Circuit Current Calculations Procedures and Methods

Per-Unit Method

11

3 Short Circuit Calculation Per-Unit Method*The per-unit method is generally used for calculating

short-circuit currents when the electrical system is more

complex.

After establishing a one-line diagram of the system,proceed to the following calculations: **

Step 1. PUXutility =KVAbase

S.C. KVAutility

Step 2. PUXtrans =(%X)(KVAbase )

(100)(KVAtrans)

PURtrans =(%R)(KVAbase)

(100)(KVAtrans)

Step 3. PUXcomponent (cable, =(X)(KVAbase)

switches, CT, bus) (1000)(KV)2

Step 4. PURcomponent (cable, =(R)( KVAbase)

switches, CT, bus) (1000)(KV)2

Step 5. Next, total all per-unit X and all per-unit R in systemto point of fault.

Step 6. Determine the per-unit impedance of the system by:

PUZtotal = (PURtotal)2 + (PUXtotal)2

Step 7. Calculate the symmetrical RMS short-circuit currentat the point of fault.

IS.C. sym RMS=KVAbase

3 (KV)(PUZtotal)

considered, such as supplied by 4 wire, 208Y/120 and

480Y/277 volt 3 phase systems, the generally accepted

procedure is to assume 50% motor load based on the full

load current rating of the transformer.)

Step 9. The symmetrical motor contribution can beapproximated by using an average multiplying factor

associated with the motors in the system. This factor varies

according to motor design and in this text may be chosen

as 4 times motor full load current for approximate

calculation purposes. To solve for the symmetrical motorcontribution:

***Isym motor contrib = (4) x (Ifull load motor)

Step 10. The total symmetrical short-circuit rms current iscalculated as:

Itotal S.C. sym RMS = (IS.C. sym RMS) + (Isym motor contrib)

Step 11. Determine X/R ratio of the system to the point offault.

X/Rratio =

PUXtotalPURtotal

Step 12. From Table 8, Column Mm, obtain the asymmetricalfactor corresponding to the X/R ratio determined in Step

11. This multiplier will provide the worst case asymmetry

occurring in the first 1/2 cycle. When the average 3-phase

multiplier is desired use column Ma.

Step 13. The asymmetrical RMS short-circuit current can

be calculated as:

IS.C. asym RMS = (IS.C. sym RMS) x (Asym Factor)

Step 14. The short-circuit current that the motor load can

contribute is an asymmetrical current usually approximated

as being equal to the locked rotor current of the motor.***

As a close approximation with a margin of safety use:

***Iasym motor contrib = (5) x (Ifull load motor)

Step 15. The total asymmetrical short-circuit RMS current

is calculated as:

ItotalS.C. asym RMS = (IS.C. asym RMS) + (Iasym motor contrib)

* The base KVA used throughout this text will be 10,000 KVA.** As in the ohmic method procedure, all ohmic values are single-phase distance one way, later compensated for in the three phase short-circuit formula by the

factor, 3. (See Step 7.) UL Listed transformers 25KVA and larger have a 10% impedance tolerance. Short circuit amperes can be affected by this tolerance. Only per-unit X is considered in this procedure since utility X/R ratio is usually quite high. For more finite details obtain per-unit R of utility source.

*** A more exact determination depends upon the sub-transient reactance of the motors in question and associated circuit impedances. A less conservativemethod would involve the total motor circuit impedance to a common bus (sometimes referred to as a zero reactance bus).

Arithmetical addition results in conservative values of fault current. More finite values involve vectorial addition of the currents.

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*

10,000 KVA Base

PUR PUX

PUX =10,000 = 0.0001 0.0001

100,000,000

PUX =(3.45) (10,000)

= 0.2300 0.2300(100) (1500)

PUR =(.56) (10,000)

= 0.0373 0.0373 (100) (1500)

(25')x

(.0379)x (10,000)

PUX =(1000) (6)

= 0.00685 0.00685(1000) (.480)2

(25') x (.0244) x (10,000)(1000) (6)

PUR =(1000) (.480)2

= 0.0044 0.0044

PUX =(.00005) (10,000)

= 0.00217 0.00217(1000) (.480)2

Total PUR and PUX = 0.0417 0.2391

PUZtotal = (0.0417)2 + (0.2391)2 = .2430

IS.C. sym RMS =10,000 = 49,489A

3 (.480)(.2430)

Isym motor contrib = 4 x 1804 = 7,216A

Itotal S.C. sym RMS = 49,489 + 7,216 = 56,705A(fault X1)

X/Rratio =.2391 = 5.73.0417

Asym Factor = 1.294 (Table 8)

IS.C. asym RMS = 49,489 x 1.294 = 64,039A

Iasym motor contrib = 5 x 1804 = 9,020A(100% motor load)

Itotal S.C. asym RMS = 64,039 + 9,020 = 73,059A(fault X1)

3 Short-Circuit Current Calculations Procedures and Methods

Per-Unit Method To Fault X1 System A

12

Available UtilityS.C. MVA 100,000

1500 KVA Transformer,480V, 3,3.5%Z, 3.45%X, .56%R

If.l. trans = 1804A

25 - 500kcmil6 Per PhaseService EntranceConductors in Steel Conduit

2000A Switch

KRP-C-2000SP Fuse

MM

Impedance DiagramOne-Line Diagram

Note: See Per Unit Method Procedure for Formulas.Actual motor contribution will be somewhat smaller than calculateddue to impedance of the feeder cable.

1 1

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M M

3 Short-Circuit Current Calculations - Procedures and Methods

Per-Unit Method To Fault X2 System A

13

Fault X1

400A Switch

LPS-RK400SP Fuse

50 - 500kcmilFeeder Cable inSteel Conduit

Motor Contribution

One-Line Diagram Impedance Diagram10,000 KVA Base

PUR PUX

PUX = .2391 .2391PUR = .0417 .0417

PUX =(.00008) (10,000)

= .0034 .0034(1000) (.480)2

50 x (.0379) x (10,000)

PUX =1000

= .0822 .0822(1000) (.480)2

50 x (.0244) x (10,000)

PUR =1000 = .0529 .0529

(1000) (.480)2

Total PUR and PUX = .0946 .3247

PUZtotal = (.0946)2

+ (.3247)2

= 0.3380

IS.C. sym RMS =10,000 = 35,621A

3 (.480)(.3380)

Isym motor contrib = 4 x 1804 = 7,216A

Itotal S.C. sym RMS = 35,621 + 7,216 = 42,837A(fault X2)

X/Rratio =.32477

= 3.43.09465

Asym Factor = 1.149 (Table 8)

IS.C. asym RMS = 1.149 x 35,621 = 40,929A

Iasym motor contrib = 5 x 1804 = 9,020A(100%motor load)

Itotal S.C. asym RMS = 40,929 + 9,020 = 49,949A(fault X2)

1

1 1

2 2

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10,000KVA BasePUR PUX

PUX =10,000 = .02 .02

500,000

PUX =(3.45) (10,000)

= .345 .345(100) (1000)

PUR =(.6) (10,000)

= .06 .06 (100) (1000)

(30')x

(.0303)

PUX =(1000) (4)

x (10,000)

= .0099 .0099(1000) (.48)2

(30') x (.0220)

PUR =(1000) (4)

x (10,000)

= .0072 .0072 (1000) (.48)2

PUX =(.00005) (10,000)

= .0022 .0022(1000) (.48)2

Total PUR and PUX = .0672 .3771

PUZtotal

= (.0672)2 + (.3771)2 = .383

IS.C. sym RMS =10,000 = 31,405A

3 (.48)(.383)

X/Rratio =.3771 = 5.62.0672

Asym Factor = 1.285 (Table 8)

IS.C.asym RMS = 31,405 x 1.285 = 40,355A

14

3 Short-Circuit Current Calculations Procedures and Methods

Per-Unit Method To Fault X1 System B

Available UtilityS.C. KVA 500,000

1000 KVA Transformer,480V, 33.45%X, .60%R

30' - 500kcmil

4 Per PhaseCopper in PVC Conduit

1600A Switch

KRP-C-1500SP Fuse

One-Line Diagram Impedance Diagram

1 1

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3 Short-Circuit Current Calculations Procedures and Methods

Per-Unit Method To Fault X2 System B

15

One-Line Diagram Impedance Diagram10,000 KVA

PUR PUX

X1 = .3771 .3771

R1 = .0672 .0672

PUX =(.00008) (10,000)

= .0035 .0035(1000) (.48)2

PUX =

(20')x

(.0327)x (10,000)

= .0142 .0142(1000) (2)

(1000) (.48)2

(20')x

(.0812)x (10,000)

PUR =(1000) (2)

= .0352 .0352 (1000) (.48)2

PUX =(.998) (10,000)

= .4435 .4435(100) (225)

PUR =(.666) (10,000)

= .296 .296 (100) (225)

Total PUR and PUX .3984 .8383

PUZtotal = (.3984)2 + (.8383)2 = .928

IS.C.sym RMS = 10,000 = 29,911A(3)(.208)(.928)

X/Rratio =.8383 = 2.10.3984

Asym Factor=

1.0491 (Table 8)IS.C. asym RMS = 29,911 x 1.0491 = 31,380A

400A Switch

LPS-RK-350SP Fuse

20 - 2/02 Per PhaseCopper in PVC conduit

225KVA Transformer,208V, 3.998%X, .666%R

2 2

1 1

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BUSSPOWER is a Computer Software Program which

calculates three phase fault currents. It is a part of the

TRON Software Package for Power Systems Analysis.

The user inputs data which includes:

- Cable and Busway Lengths and Types- Transformer Rating and Impedence- Fault sources such as Utility Available and Motor

Contribution.

Following the data input phase, the program is

executed and an output report reviewed.

The following is a partial output report of System A

being studied.

TRON Software Fault Calculation Program Three Phase Fault Report

SYSTEM AFault Study Summary

Bus Record Voltage Available RMS Duties

Name L-L 3 Phase Momentary

(Sym) (Asym)

X1 480 58414 77308

X2 480 44847 53111

The following is a partial output report of the

distribution System B.

SYSTEM BFault Study Summary

Bus Record Voltage Available RMS Duties

Name L-L 3 Phase Momentary

(Sym) (Asym)X1 480 31,363 40,141

X2 208 29,980 31,425

A further description of this program and its

capabilities is on the back cover of this bulletin.

3 Short-Circuit Current Calculations Procedures and Methods

TRON Computer Software Method

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The application of the point-to-point method permits the

determination of available short-circuit currents with a

reasonable degree of accuracy at various points for either

3 or 1 electrical distribution systems. This method can

assume unlimited primary short-circuit current (infinite bus).

Basic Point-to-Point Calculation ProcedureStep 1. Determine the transformer full load amperes fromeither the nameplate or the following formulas:

3 Transformer If.l. =KVA x 1000EL-L x 1.732

1 Transformer If.l. =KVA x 1000

EL-L

Step 2. Find the transformer multiplier.

Multiplier = 100*%Z trans

Note. Transformer impedance (Z) helps to determine what the short circuitcurrent will be at the transformer secondary. Transformer impedance is

determined as follows: The transformer secondary is short circuited. Voltageis applied to the primary which causes full load current to flow in thesecondary. This applied voltage divided by the rated primary voltage is theimpedance of the transformer.Example: For a 480 volt rated primary, if 9.6 volts causes secondary full loadcurrent to flow through the shorted secondary, the transformer impedance is9.6/480 = .02 = 2%Z.In addition, UL listed transformer 25KVA and larger have a 10%impedance tolerance. Short circuit amperes can be affected by thistolerance.

Step 3. Determine the transformer let-thru short-circuitcurrent**.

IS.C. = If.l. x Multiplier

Note. Motor short-circuit contribution, if significant, may be added to the

transformer secondary short-circuit current value as determined in Step 3.Proceed with this adjusted figure through Steps 4, 5 and 6. A practicalestimate of motor short-circuit contribution is to multiply the total motorcurrent in amperes by 4.

Step 4. Calculate the "f" factor.

3 Faults f = 1.732 x L x IC x EL-L

1 Line-to-Line (L-L)2 x L x I

Faults on 1 Center f =C x EL-LTapped Transformer

1 Line-to-Neutral2 x L x I(L-N) Faults on 1 f =C x EL-NCenter Tapped Transformer

Where:L = length (feet) of circuit to the fault.C = constant from Table 6, page 27. For parallel

runs, multiply C values by the number of

conductors per phase.

I = available short-circuit current in amperes atbeginning of circuit.

Note. The L-N fault current is higher than the L-L fault current at thesecondary terminals of a single-phase center-tapped transformer. Theshort-circuit current available (I) for this case in Step 4 should be adjustedat the transformer terminals as follows:At L-N center tapped transformer terminals,I = 1.5 x L-L Short-Circuit Amperes at Transformer Terminals

At some distance from the terminals, depending upon wire size, the L-N faultcurrent is lower than the L-L fault current. The 1.5 multiplier is anapproximation and will theoretically vary from 1.33 to 1.67. These figures arebased on change in turns ratio between primary and secondary, infinitesource available, zero feet from terminals of transformer, and 1.2 x %X and1.5 x %R for L-N vs. L-L resistance and reactance values. Begin L-Ncalculations at transformer secondary terminals, then proceed point-to-point.

Step 5. Calculate "M" (multiplier).

M = 11 + f

Step 6. Calculate the available short-circuit symmetricalRMS current at the point of fault.

IS.C. sym RMS = IS.C. x M

Calculation of Short-Circuit Currentsat Second Transformer in System

Use the following procedure to calculate the level of

fault current at the secondary of a second, downstream

transformer in a system when the level of fault current at the

transformer primary is known.

Procedure for Second Transformer in System

Step 1. Calculate the "f" factor (IS.C. primary known)

3 Transformer(IS.C. primary and f =

IS.C. primary x Vprimary x 1.73 (%Z)

IS.C. secondary are 100,000 x KVA trans3 fault values)

1 Transformer(IS.C. primary andIS.C. secondary are f =

IS.C. primary x Vprimary x (%Z)

1 fault values: 100,000x KVA trans

IS.C. secondary is L-L)

Step 2. Calculate "M" (multiplier).

M =1

1 + f

Step 3. Calculate the short-circuit current at the secondaryof the transformer. (See Note under Step 3 of "Basic Point-

to-Point Calculation Procedure".)

IS.C. secondary =Vprimary x M x IS.C. primary

Vsecondary

**

*

3 Short-Circuit Current Calculations Procedures and Methods

Point-to-Point Method

17

IS.C. primary

MAINTRANSFORMER

H.V. UTILITYCONNECTION

IS.C. secondary

IS.C. primary IS.C. secondary

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Available UtilityS.C. MVA 100,000

1500 KVA Transformer,480V, 3, 3.5%Z,3.45%X, 56%R

If.l. =1804A

25' - 500kcmil6 Per PhaseService EntranceConductors in Steel Conduit

2000A Switch

KRP-C-2000SP Fuse

Fault X1

400A Switch

LPS-RK-400SP Fuse

50' - 500 kcmilFeeder Cablein Steel Conduit

Fault X2

Motor Contribution M

3 Short-Circuit Current Calculations Procedures and Methods

Point-to-Point Method To Faults X1 & X2 System A

Fault X1

Step 1. If.l. =1500 x 1000 = 1804A480 x 1.732

Step 2. Multiplier = 100 = 28.573.5

Step 3. IS.C.= 1804 x 28.57 = 51,540A

Step 4. f =1.732 x 25 x 51,540 = 0.0349

6 x 22,185 x 480

Step 5. M =1 = .9663

1 + .0349

Step 6. IS.C.sym RMS = 51,540 x .9663 = 49,803A

IS.C.motor contrib = 4 x 1,804 = 7,216A

ItotalS.C. sym RMS = 49,803 + 7,216 = 57,019A(fault X1)

Fault X2

Step 4. Use IS.C.sym RMS @ Fault X1 to calculate "f"

f =1.732 x 50 x 49,803 = .4050

22,185 x 480

Step 5. M =1 = .7117

1 + .4050

Step 6. IS.C.sym RMS = 49,803 x .7117 = 35,445A

Isym motor contrib = 4 x 1,804 = 7,216A

Itotal S.C. sym RMS(fault X2)

= 35,445 + 7,216 = 42,661A

1

2

18

One-Line Diagram

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3 Short-Circuit Current Calculations RMS Amperes

Comparison of Results

Fault X1

Step 1. If.l. =1000 x 1000 = 1203A480 x 1.732

Step 2. Multiplier = 100 = 28.573.5

Step 3. IS.C. = 1203 x 28.57 = 34,370A

Step 4. f =1.732 x 30 x 34,370 = .0348

4 x 26,706 x 480

Step 5. M =1 = .9664

1 + .0348

Step 6. IS.C.sym RMS = 34,370 x .9664 = 33,215A

Fault X2

Step 4. f = 1.732 x 20 x 33,215 = .10492 x 11,423 x 480

Step 5. M = 1 = .9051 + .1049

Step 6. IS.C.sym RMS = 33,215 x .905 = 30,059A

Fault X2

f =30,059 x 480 x 1.732 x 1.2 = 1.333

100,000 x 225

M =1 = .4286

1 + 1.333

IS.C. sym RMS =480 x .4286 x 30,059 = 29,731A

208

3 Short-Circuit Current Calculations Procedures and Methods

Point-to-Point Method To Faults X1 & X2 - System B

Available Utility500,000 S.C KVA

1000 KVA Transformer,480V, 3,3.5%Z

If.l.= 1203A

30 - 500 kcmil4 Per PhaseCopper in PVC Conduit

1600A Switch

KRP-C-1500SP Fuse

Fault X1

400A Switch

LPS-RK-350SP Fuse

20 - 2/02 Per PhaseCopper in PVC Conduit

225 KVA transformer,208V, 31.2%Z

One-Line Diagram

1

2

19

System A

Ohmic Per-Unit TRON PTPSym. Asym. Sym. Asym. Sym. Asym. Sym.

X1W/O Motor 49,489 64,039 49,489 64,039 49,992 64,430 49,803

W/Motor 56,705 73,059 56,705 73,059 58,414 77,308 57,019

X2W/O Motor 35,621 40,929 35,621 40,929 36,126 41,349 35,445

W/Motor 42,837 49,949 42,837 49,949 44,847 53,111 42,661

Notes:1. OHMIC and PER UNIT methods assume 100% motor contribution at X1,

then at X2.2. TRON modeled 100% motor contribution by assuming 1500 HP load,

located at Point X2.3. PTP method added symmetrical motor contribution at X1, then at X2.

System B

Ohmic Per-Unit TRON PTPSym. Asym. Sym. Asym. Sym. Asym. Sym.

X1 31,405 40,355 31,405 40,355 31,363 40,145 33,215

X2 29,911 31,380 29,911 31,380 29,980 31,425 29,731

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1 Short-Circuit Current Calculations 1 Transformer System

Procedures and Methods

Short-circuit calculations on a single-phase center

tapped transformer system require a slightly different

procedure than 3 faults on 3 systems.

1. It is necessary that the proper impedance be used to

represent the primary system. For 3 fault calculations, asingle primary conductor impedance is only considered

from the source to the transformer connection. This is

compensated for in the 3 short-circuit formula by

multiplying the single conductor or single-phase

impedance by 1.73.

However, for single-phase faults, a primary conductor

impedance is considered from the source to the

transformer and back to the source. This is compensated in

the calculations by multiplying the 3 primary source

impedance by two.

2. The impedance of the center-tapped transformer must

be adjusted for the half-winding (generally line-to-neutral)fault condition.

The diagram at the right illustrates that during line-to-

neutral faults, the full primary winding is involved but, only

the half-winding on the secondary is involved. Therefore,

the actual transformer reactance and resistance of the half-

winding condition is different than the actual transformer

reactance and resistance of the full winding condition.

considering line-to-neutral faults. The adjustment multipliers

generally used for this condition are as follows:

1.5 times full winding %R on full winding basis.

1.2 times full winding %X on full winding basis.

Note: %R and %X multipliers given in Table 1.3 may be used, however,calculatios must be adjusted to indicate transformer KVA/2.

3. The impedance of the cable and two-pole switches on

the system must be considered "both-ways" since the

current flows to the fault and then returns to the source. For

instance, if a line-to-line fault occurs 50 feet from a

transformer, then 100 feet of cable impedance must be

included in the calculation.

The calculations on the following pages illustrate 1

fault calculations on a single-phase transformer system.

Both line-to-line and line-to-neutral faults are considered.

Note in these examples:

a. The multiplier of 2 for some electrical components to

account for the single-phase fault current flow,

b. The half-winding transformer %X and %R multipliers for

the line-to-neutral fault situation,and

c. The KVA and voltage bases used in the per-unit

calculations

PRIMARY

SECONDARY

SHORTCIRCUIT

PRIMARY

SECONDARY

SHORT CIRCUIT

L2 N L1

SHORT CIRCUIT

L1

N

L2

A

B

C

50 feet

20

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100,000 KVA3 Source

75KVA, 1 Transformer,1.22%X, .68%R

Negligible Distance

400A Switch

LPN-RK-400SP Fuse

25' - 500kcmil

Magnetic Conduit

1 Short-Circuit Current Calculations 1 Transformer System

Per-Unit Method Line-to-Line Fault @ 240V Fault X1

10,000KVA Base

PUR PUX

PUX(3) =10,000 = .1100,000

PUX(1) = 2 x .1 = .2000 .2000

PUX =(1.22) (10,000)

= 1.6267 1.6267(100) (75)

PUR =(.68) (10,000)

= .9067 .9067 (100) (75)

PUX =2(.00008) (10,000)

= .0278 .0278(1000) (.240)2

25'

PUX =2 x

1000x .0379 x 10,000

= .3289 .3289(1000) (.240)2

25'

PUR=2 x

1000x .0244 x 10,000

= .2118 .2118 (1000) (.240)2

Total PUR and PUX = 1.1185 2.1834

PUZtotal = (1.1185)2 + (2.1834)2 = 2.4532

IS.C. sym RMS =10,000

= 16,984AL-L @ 240V (.240) (2.4532)

Note: See "Data Section" for impedance data for the electrical components.

One-Line Diagram

1 1

21

Impedance Diagram

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*

**

1 Short-Circuit Current Calculations 1 Transformer System

Per-Unit Method Line-to-Neutral Fault @ 120V Fault X1

100,000 KVA3 Source

75KVA, 1 Transformer,1.22%X, .68%R

Negligible Distance

400A Switch

LPN-RK-400SP Fuse

25' - 500kcmil

Magnetic Conduit

10,000KVA Base

PUR PUX

PUX(3) =10,000 = .1100,000

PUX(1) = 2 x .1 = .2000 .2000

PUX =(1.2) (1.22) (10,000)

= 1.952 1.952(100) (75)

PUR =(1.5) (.68) (10,000)

= 1.3600 1.3600 (100) (75)

PUX* =(.00008) (10,000)

= .0556 .0556(1000) (.120)2

25'

PUX** =2 x

1000x .0379 x 10,000

= 1.316 1.316(1000) (.120)2

25'

PUR** =2 x

1000x .0244 x 10,000

= .8472 .8472 (1000) (.120)2

Total PUR and PUX = 2.2072 3.5236

PUZtotal = (2.2072)2 + (3.5236)2 = 4.158

IS.C. sym RMS =10,000

= 20,041AL-N @ 120V (.120) (4.158)

Note: See "Data Section" for impedance data for the electrical components.

The multiplier of two (2) is not applicable since on a line to neutral fault, onlyone switch pole is involved.

Assumes the neutral conductor and the line conductor are the same size.

1 1

One-Line Diagram Impedance Diagram

22

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1 Short-Circuit Current Calculations 1 Transformer System

Point-to-Point Method Line-to-Line Fault @ 240V Fault X1

Available UtilityS.C.KVA 100,0003 Source

75KVA, 1 Transformer,1.22%X, .68%R1.40%Z120/240V

Negligible Distance

400A Switch

LPN-RK-400SP Fuse

25' - 500kcmil

Magnetic Conduit

One-Line Diagram

1

Fault X1

Step 1. If.l. =75 x 1000 = 312.5A

240

Step 2. Multiplier =100 = 71.431.40

Step 3. IS.C. = 312.5 x 71.43 = 22,322A

Step 4. f = 2x 25 x 22,322 = .2096

22,185 x 240

Step 5.1 = .8267M =

1 + .2096

Step 6. IS.C. L-L (X1) = 22,322 x .8267 = 18,453A

23

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*

1 Short Circuit Calculations RMS Amperes

Comparison of Results

Per-Unit Method vs. Point-to-Point MethodPer-Unit PTP

Method Method

X1Line-Line 16,984A 18,453A

Line-Neutral 20,041A 20,555A

Fault X1

Step 1. If.l. =75 x 1000 = 312.5A

240

Step 2. Multiplier =100 = 71.431.40

Step 3. IS.C. (L-L) = 312.5 x 71.43 = 22,322A

IS.C. (L-N) = 22,322 x 1.5 = 33,483A

Step 4. f =2*x 25 x 22,322 x 1.5

= .628822,185 x 120

Step 5. 1 = .6139M =1 + .6288

Step 6. IS.C. L-N (X1) = 33,483 x .6139 = 20,555A

Assumes the Neutral conductor and the line conductor are the same size.

Available UtilityS.C.KVA 100,0003 Source

75KVA, 1 Transformer,1.22% X, .68%R,1.40%Z120/240V

Negligible Distance

400A Switch

LPN-RK-400SP Fuse

25' - 500kcmil

Magnetic Conduit

One-Line Diagram

1 Short-Circuit Current Calculations 1 Transformer System

Point-to-Point Method Line-to-Neutral Fault @ 120V Fault X1

1

24

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*

**

Table 1.4. Impedance Data for Single Phase and Three PhaseTransformers-Supplement

KVA Suggested

1 3 %Z X/R Ratio for Calculation

10 1.2 1.1

15 1.3 1.175 1.11 1.5

150 1.07 1.5

225 1.12 1.5

300 1.11 1.5

333 1.9 4.7

500 2.1 5.5

These represent actual transformer nameplate ratings taken from field

installations.

Note: UL Listed transformers 25KVA and greater have a 10% tolerance on

their impedance nameplate.

Table 2. Current Transformer Reactance DataApproximate Reactance of Current Transformers*

Reactance in Ohms for

Primary Current Various Voltage Ratings

Ratings - Amperes 600-5000V 7500V 15,000V

100 - 200 0.0022 0.0040

250 - 400 0.0005 0.0008 0.0002

500 - 800 0.00019 0.00031 0.00007

1000 - 4000 0.00007 0.00007 0.00007

Note: Values given are in ohms per phase. For actual values, refer to manu-

facturers' data.

Practice for Commercial Building Power Systems, Copyright 1990 by the

Institute of Electrical and Electronics Engineers, Inc. with the permission of

the IEEE Standards Department.

Table 3. Disconnecting Switch Reactance Data(Disconnecting-Switch Approximate Reactance Data, in Ohms*)Switch Size Reactance(Amperes) (Ohms)

200 0.0001400 0.00008

600 0.00008

800 0.00007

1200 0.00007

1600 0.00005

2000 0.00005

3000 0.00004

4000 0.00004

Note: The reactance of disconnecting switches for low-voltage circuits

(600V and below) is in the order of magnitude of 0.00008 - 0.00005

ohm/pole at 60 Hz for switches rated 400 - 4000 A, respectively.

*For actual values, refer to manufacturers data.

Practice for Commercial Building Power Systems, Copyright 1990 by the

Institute of Electrical and Electronics Engineers, Inc. with the permission ofthe IEEE Standards Department.

1 Pole

25

Table 1.1. Transformer Impedance Data(X/R Ratio of Transformers Based on ANSI/IEEE C37.010-1979)

Practice for Electric Power Distribution for Industrial Plants, Copyright 1986

by the Institute of Electrical and Electronics Engineers, Inc with the

permission of the IEEE Standards Department.

Table 1.2. Impedance Data for Three Phase TransformersKVA %R %X %Z X/R

3.0 3.7600 1.0000 3.8907 0.265

6.0 2.7200 1.7200 3.2182 0.632

9.0 2.3100 1.1600 2.5849 0.502

15.0 2.1000 1.8200 2.7789 0.867

30.0 0.8876 1.3312 1.6000 1.5

45.0 0.9429 1.4145 1.7000 1.5

75.0 0.8876 1.3312 1.6000 1.5

112.5 0.5547 0.8321 1.0000 1.5

150.0 0.6657 0.9985 1.2000 1.5

225.0 0.6657 0.9985 1.2000 1.5

300.0 0.6657 0.9985 1.2000 1.5

500.0 0.7211 1.0816 1.3000 1.5

750.0 0.6317 3.4425 3.5000 5.451000.0 0.6048 3.4474 3.5000 5.70

1500.0 0.5617 3.4546 3.5000 6.15

2000.0 0.7457 4.9441 5.0000 6.63

2500.0 0.7457 4.9441 5.0000 6.63

Note: UL Listed transformers 25KVA and greater have a 10% tolerance on

their nameplate impedance.

Table 1.3. Impedance Data for Single Phase TransformersSuggested Normal Range Impedance Multipliers**

X/R Ratio of Percent For Line-to-Neutral

kVA for Impedance (%Z)* Faults

1 Calculation for %X for%R

25.0 1.1 1.26.0 0.6 0.75

37.5 1.4 1.26.5 0.6 0.75

50.0 1.6 1.26.4 0.6 0.75

75.0 1.8 1.26.6 0.6 0.75

100.0 2.0 1.35.7 0.6 0.75

167.0 2.5 1.46.1 1.0 0.75

250.0 3.6 1.96.8 1.0 0.75

333.0 4.7 2.46.0 1.0 0.75

500.0 5.5 2.25.4 1.0 0.75

National standards do not speciify %Z for single-phase transformers. Consultmanufacturer for values to use in calculation.Based on rated current of the winding (onehalf nameplate kVA divided bysecondary line-to-neutral voltage).

Note: UL Listed transformers 25 KVA and greater have a 10% tolerance ontheir impedance nameplate.

Recommended Practice for Protection and Coordination of Industrial and

Commercial Power Systems, Copyright 1986 by the Institute of Electrical

and Electronics Engineers, Inc. with the permission of the IEEE Standards

Department.

50

40

30

20

10

0

TypicalX/R

0.5 1 2 5 10 20 50 100 200 500 1000Self-Cooled Transformer Rating in MVA

Data Section

Impedance and Reactance DataTransformers and Switches

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Data Section

Impedance & Reactance Data-Circuit Breakers and Conductors

Table 5. Impedance Data - Insulated Conductors(Ohms/1000 ft. each conductor - 60Hz)Size Resistance (25C) Reactance - 600V - THHN

AWG or Copper Aluminum Single Conductors 1 Multiconductor

kcM Metal NonMet Metal Nonmet Mag. Nonmag. Mag Nonmag.

14 2.5700 2.5700 4.2200 4.2200 .0493 .0394 .0351 .030512 1.6200 1.6200 2.6600 2.6600 .0468 .0374 .0333 .0290

10 1.0180 1.0180 1.6700 1.6700 .0463 .0371 .0337 .0293

8 .6404 .6404 1.0500 1.0500 .0475 .0380 .0351 .0305

6 .4100 .4100 .6740 .6740 .0437 .0349 .0324 .0282

4 .2590 .2590 .4240 .4240 .0441 .0353 .0328 .0235

2 .1640 .1620 .2660 .2660 .0420 .0336 .0313 .0273

1 .1303 .1290 .2110 .2110 .0427 .0342 .0319 .0277

1/0 .1040 .1020 .1680 .1680 .0417 .0334 .0312 .0272

2/0 .0835 .0812 .1330 .1330 .0409 .0327 .0306 .0266

3/0 .0668 .0643 .1060 .1050 .0400 .0320 .0300 .0261

4/0 .0534 .0511 .0844 .0838 .0393 .0314 .0295 .0257

250 .0457 .0433 .0722 .0709 .0399 .0319 .0299 .0261

300 .0385 .0362 .0602 .0592 .0393 .0314 .0295 .0257

350 .0333 .0311 .0520 .0507 .0383 .0311 .0290 .0254

400 .0297 .0273 .0460 .0444 .0385 .0308 .0286 .0252

500 .0244 .0220 .0375 .0356 .0379 .0303 .0279 .0249

600 .0209 .0185 .0319 .0298 .0382 .0305 .0278 .0250750 .0174 .0185 .0264 .0240 .0376 .0301 .0271 .0247

1000 .0140 .0115 .0211 .0182 .0370 .0296 .0260 .0243

Note: Increased resistance of conductors in magnetic raceway is due to the effect of hysteresislosses. The increased resistance of conductors in metal non-magnetic raceway is due to the effectof eddy current losses. The effect is essentially equal for steel and aluminum raceway. Resistancevalues are acceptable for 600 volt, 5KV and 15 KV insulated Conductors.

Size Reactance - 5KV Reactance - 15KV

AWG or Single Conductors 1 Multiconductor Single Conductors 1 Multiconductor

kcM Mag. Nonmag. Mag. Nonmag. Mag. Nonmag. Mag. Nonmag.

8 .0733 .0586 .0479 .0417

6 .0681 .0545 .0447 .0389 .0842 .0674 .0584 .0508

4 .0633 .0507 .0418 .0364 .0783 .0626 .0543 .0472

2 .0591 .0472 .0393 .0364 .0727 .0582 .0505 .0439

1 .0571 .0457 .0382 .0332 .0701 .0561 .0487 .0424

1/0 .0537 .0430 .0360 .0313 .0701 .0561 .0487 .0424

2/0 .0539 .0431 .0350 .0305 .0661 .0561 .0458 .0399

3/0 .0521 .0417 .0341 .0297 .0614 .0529 .0427 .03724/0 .0505 .0404 .0333 .0290 .0592 .0491 .0413 .0359

250 .0490 .0392 .0323 .0282 .0573 .0474 .0400 .0348

300 .0478 .0383 .0317 .0277 .0557 .0458 .0387 .0339

350 .0469 .0375 .0312 .0274 .0544 .0446 .0379 .0332

400 .0461 .0369 .0308 .0270 .0534 .0436 .0371 .0326

500 .0461 .0369 . 0308 .0270 .0517 .0414 .0357 .0317

600 .0439 .0351 .0296 .0261 .0516 .0414 . 0343 .0309

750 .0434 .0347 .0284 .0260 .0500 .0413 .0328 .0301

1000 .0421 .0337 .0272 .0255 .0487 .0385 .0311 .0291

These are only representative figures. Reactance is affected by cable insulation type, shielding,

conductor outside diameter, conductor spacing in 3 conductor cable, etc. In commercial buildings

meduim voltage impedances normally do not affect the short circuit calculations significantly.

This table has been reprinted from IEEE Std 241-1990, IEEE Recommended Practice for Commercial

Building Power Systems, copyright 1990 by the Institute of Electrical and Electronics Engineers,

Inc. with the permission of the IEEE Standards Department.

*

Table 4. Circuit Breaker Reactance Data(a) Reactance of Low-Voltage Power Circuit Breakers

Circuit-Breaker

Interrupting Circuit-Breaker

Rating Rating Reactance(amperes) (amperes) (ohms)

15,000 15 - 35 0.04

and 50 - 100 0.004

25,000 125 - 225 0.001

250 - 600 0.0002

50,000 200 - 800 0.0002

1000 - 1600 0.00007

75,000 2000 - 3000 0.00008

100,000 4000 0.00008

(b)Typical Molded Case Circuit Breaker Impedances

Molded-Case

Circuit-Breaker

Rating Resistance Reactance

(amperes) (ohms) (ohms)

20 0.00700 Negligible

40 0.00240 Negligible

100 0.00200 0.00070

225 0.00035 0.00020

400 0.00031 0.00039

600 0.00007 0.00017

Notes:

(1) Due to the method of rating low-voltage powercircuit breakers, the reactance of the circuit breakerwhich is to interrupt the fault is not included incalculating fault current.

(2) Above 600 amperes the reactance of molded casecircuit breakers are similar to those given in (a)

For actual values, refer to manufacturers data.

IEEE Recommended Practice for Commercial Building

Power Systems, copyright 1990 by the Institute of

Electrical and Electronics Engineers, Inc. with the

permission of the IEEE Standards Department.

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Data Section

"C" Values for Conductors and Busway

Table 6. C Values for Conductors and BuswayCopper

AWG Three Single Conductors Three-Conductor Cable

or Conduit Conduit

kcmil Steel Nonmagnetic Steel Nonmagnetic

600V 5KV 15KV 600V 5KV 15KV 600V 5KV 15KV 600V 5KV 15KV

14 389 389 389 389 389 389 389 389 389 389 389 389

12 617 617 617 617 617 617 617 617 617 617 617 617

10 981 981 981 981 981 981 981 981 981 981 981 981

8 1557 1551 1557 1558 1555 1558 1559 1557 1559 1559 1558 1559

6 2425 2406 2389 2430 2417 2406 2431 2424 2414 2433 2428 2420

4 3806 3750 3695 3825 3789 3752 3830 3811 3778 3837 3823 3798

3 4760 4760 4760 4802 4802 4802 4760 4790 4760 4802 4802 4802

2 5906 5736 5574 6044 5926 5809 5989 5929 5827 6087 6022 5957

1 7292 7029 6758 7493 7306 7108 7454 7364 7188 7579 7507 7364

1/0 8924 8543 7973 9317 9033 8590 9209 9086 8707 9472 9372 9052

2/0 10755 10061 9389 11423 10877 10318 11244 11045 10500 11703 11528 11052

3/0 12843 11804 11021 13923 13048 12360 13656 13333 12613 14410 14118 13461

4/0 15082 13605 12542 16673 15351 14347 16391 15890 14813 17482 17019 16012

250 16483 14924 13643 18593 17120 15865 18310 17850 16465 19779 19352 18001

300 18176 16292 14768 20867 18975 17408 20617 20051 18318 22524 21938 20163

350 19703 17385 15678 22736 20526 18672 19557 21914 19821 22736 24126 21982

400 20565 18235 16365 24296 21786 19731 24253 23371 21042 26915 26044 23517

500 22185 19172 17492 26706 23277 21329 26980 25449 23125 30028 28712 25916

600 22965 20567 47962 28033 25203 22097 28752 27974 24896 32236 31258 27766

750 24136 21386 18888 28303 25430 22690 31050 30024 26932 32404 31338 28303

1000 25278 22539 19923 31490 28083 24887 33864 32688 29320 37197 35748 31959

Aluminum

14 236 236 236 236 236 236 236 236 236 236 236 236

12 375 375 375 375 375 375 375 375 375 375 375 375

10 598 598 598 598 598 598 598 598 598 598 598 598

8 951 950 951 951 950 951 951 951 951 951 951 951

6 1480 1476 1472 1481 1478 1476 1481 1480 1478 1482 1481 1479

4 2345 2332 2319 2350 2341 2333 2351 2347 2339 2353 2349 2344

3 2948 2948 2948 2958 2958 2958 2948 2956 2948 2958 2958 2958

2 3713 3669 3626 3729 3701 3672 3733 3719 3693 3739 3724 3709

1 4645 4574 4497 4678 4631 4580 4686 4663 4617 4699 4681 4646

1/0 5777 5669 5493 5838 5766 5645 5852 5820 5717 5875 5851 5771

2/0 7186 6968 6733 7301 7152 6986 7327 7271 7109 7372 7328 7201

3/0 8826 8466 8163 9110 8851 8627 9077 8980 8750 9242 9164 8977

4/0 10740 10167 9700 11174 10749 10386 11184 11021 10642 11408 11277 10968

250 12122 11460 10848 12862 12343 11847 12796 12636 12115 13236 13105 12661

300 13909 13009 12192 14922 14182 13491 14916 14698 13973 15494 15299 14658

350 15484 14280 13288 16812 15857 14954 15413 16490 15540 16812 17351 16500

400 16670 15355 14188 18505 17321 16233 18461 18063 16921 19587 19243 18154

500 18755 16827 15657 21390 19503 18314 21394 20606 19314 22987 22381 20978

600 20093 18427 16484 23451 21718 19635 23633 23195 21348 25750 25243 23294

750 21766 19685 17686 23491 21769 19976 26431 25789 23750 25682 25141 23491

1000 23477 21235 19005 28778 26109 23482 29864 29049 26608 32938 31919 29135

Note: These values are equal to one over the impedance per foot for impedances found in Table 5, Page 26.

Ampacity Busway

Plug-In Feeder High Impedance

Copper Aluminum Copper Aluminum Copper

225 28700 23000 18700 12000

400 38900 34700 23900 21300

600 41000 38300 36500 31300

800 46100 57500 49300 44100

1000 69400 89300 62900 56200 15600

1200 94300 97100 76900 69900 16100

1350 119000 104200 90100 84000 17500

1600 129900 120500 101000 90900 19200

2000 142900 135100 134200 125000 20400

2500 143800 156300 180500 166700 21700

3000 144900 175400 204100 188700 23800

4000 277800 256400

Note: These values are equal to one over the impedance per foot for

impedances in Table 7, Page 28.

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Data Section

Busway Impedance Data

Table 7. Busway Impedance Data (Ohms per 1000 Feet Line-to-Neutral, 60 Cycles)Plug-In Busway

Copper Bus Bars Aluminum Bus Bars

Ampere Rating Resistance Reactance Impedance Resistance Reactance Impedance

225 0.0262 0.0229 0.0348 0.0398 0.0173 0.0434

400 0.0136 0.0218 0.0257 0.0189 0.0216 0.0288

600 0.0113 0.0216 0.0244 0.0179 0.0190 0.0261

800 0.0105 0.0190 0.0217 0.0120 0.0126 0.0174

1000 0.0071 0.0126 0.0144 0.0080 0.0080 0.0112

1200 0.0055 0.0091 0.0106 0.0072 0.0074 0.0103

1350 0.0040 0.0072 0.0084 0.0065 0.0070 0.0096

1600 0.0036 0.0068 0.0077 0.0055 0.0062 0.0083

2000 0.0033 0.0062 0.0070 0.0054 0.0049 0.0074

2500 0.0032 0.0062 0.0070 0.0054 0.0034 0.0064

3000 0.0031 0.0062 0.0069 0.0054 0.0018 0.0057

4000 0.0030 0.0062 0.0069

5000 0.0020 0.0039 0.0044

Low-Impedance Feeder Busway

225 0.0425 0.0323 0.0534 0.0767 0.0323 0.0832

400 0.0291 0.0301 0.0419 0.0378 0.0280 0.0470

600 0.0215 0.0170 0.0274 0.0305 0.0099 0.0320

800 0.0178 0.0099 0.0203 0.0212 0.0081 0.0227

1000 0.0136 0.0082 0.0159 0.0166 0.0065 0.0178

1200 0.0110 0.0070 0.0130 0.0133 0.0053 0.0143

1350 0.0090 0.0065 0.0111 0.0110 0.0045 0.0119

1600 0.0083 0.0053 0.0099 0.0105 0.0034 0.0110

2000 0.0067 0.0032 0.0074 0.0075 0.0031 0.0080

2500 0.0045 0.0032 0.0055 0.0055 0.0023 0.0060

3000 0.0041 0.0027 0.0049 0.0049 0.0020 0.0053

4000 0.0030 0.0020 0.0036 0.0036 0.0015 0.0039

5000 0.0023 0.0015 0.0027

The above data represents values which are a composite of those obtained by a survey of industry; values tend to be on the low side.

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Data Section

Asymmetrical Factors

Table 8. Asymmetrical FactorsRatio to Symmetrical RMS Amperes

Short Circuit Short Maximum 1 phase Maximum 1 phase Average 3 phase

Power Factor, Circuit Instantaneous RMS Amperes at RMS Amperes at

Percent* X/R Ratio Peak Amperes Mp 1/2 Cycle Mm 1/2 Cycle Ma*

(Asym.Factor)*

0 2.828 1.732 1.394

1 100.00 2.785 1.697 1.374

2 49.993 2.743 1.662 1.354

3 33.322 2.702 1.630 1.336

4 24.979 2.663 1.599 1.318

5 19.974 2.625 1.569 1.302

6 16.623 2.589 1.540 1.286

7 14.251 2.554 1.512 1.271

8 13.460 2.520 1.486 1.256

9 11.066 2.487 1.461 1.242

10 9.9301 2.455 1.437 1.229

11 9.0354 2.424 1.413 1.216

12 8.2733 2.394 1.391 1.204

13 7.6271 2.364 1.370 1.193

14 7.0721 2.336 1.350 1.182

15 6.5912 2.309 1.331 1.172

16 6.1695 2.282 1.312 1.162

17 5.7947 2.256 1.295 1.152

18 5.4649 2.231 1.278 1.144

19 5.16672 2.207 1.278 1.135

20 4.8990 2.183 1.247 1.127

21 4.6557 2.160 1.232 1.119

22 4.4341 2.138 1.219 1.112

23 4.2313 2.110 1.205 1.105

24 4.0450 2.095 1.193 1.099

25 3.8730 2.074 1.181 1.092

26 3.7138 2.054 1.170 1.087

27 3.5661 2.034 1.159 1.081

28 3.4286 2.015 1.149 1.076

29 3.3001 1.996 1.139 1.071

30 3.1798 1.978 1.130 1.064

31 3.0669 1.960 1.122 1.062

32 2.9608 1.943 1.113 1.057

33 2.8606 1.926 1.106 1.057

34 2.7660 1.910 1.098 1.050

35 2.6764 1.894 1.091 1.046

36 2.5916 1.878 1.085 1.043

37 2.5109 1.863 1.079 1.040

38 2.4341 1.848 1.073 1.037

39 2.3611 1.833 1.068 1.034

40 2.2913 1.819 1.062 1.031

41 2.2246 1.805 1.058 1.029

42 2.1608 1.791 1.053 1.027

43 2.0996 1.778 1.049 1.024

44 2.0409 1.765 1.045 1.023

45 1.9845 1.753 1.041 1.021

46 1.9303 1.740 1.038 1.019

47 1.8780 1.728 1.035 1.017

48 1.8277 1.716 1.032 1.016

49 1.7791 1.705 1.029 1.014

50 1.7321 1.694 1.026 1.013

55 1.5185 1.641 1.016 1.008

60 1.3333 1.594 1.009 1.004

65 1.1691 1.517 1.005 1.001

70 1.0202 1.517 1.002 1.001

75 0.8819 1.486 1.0008 1.0004

80 0.7500 1.460 1.0002 1.0001

85 0.6198 1.439 1.00004 1.00002

100 0.0000 1.414 1.00000 1.00000

*Reprinted by permission of National Electrical Manufacturer's Association from

NEMA Publication AB-1, 1986, copyright 1986 by NEMA.

Selective Coordination (Blackout Prevention)Having determined the faults that must be

interrupted, the next step is to specify Protective

Devices that will provide a Selectively Coordinated

System with proper Interrupting Ratings.

Such a system assures safety and reliability

under all service conditions and prevents needless

interruption of service on circuits other than the one

on which a fault occurs.

The topic of Selectivity will be Discussed in the

next Handbook, EDP II.

Component Protection (Equipment Damage Prevention)Proper protection of electrical equipment

requires that fault current levels be known. The

characteristics and let-through values of the

overcurrent device must be known, and compared

to the equipment withstand ratings. This topic ofComponent Protection is discussed in the third

Handbook, EDP III.